I've recently came across a problem that demanded to find the value of all the $x_i$ such that $P=x_1 x_2 x_3 \cdots x_n$ is Maximal, given $S=x_1+x_2+\ldots+x_n=a$ where $a$ is some natural number.

The first step is to prove that the maximal value of $P$ is achieved when all the $x_i$s are equal, which is easy to see via the arithmetico-geometric inequality, but i went ahead and tried to prove it in another way.

Let's start with the case when $n = 2$, let $K= \frac{x_1 + x_2}{2}$ so we get $x_1=K+b$ and $x_2=K-b$ for some $b$, so $P=K^2 - b^2$. In order for $P$ to be maximal, $b$ needs to be $0$, so $x_1=x_2$.

Let's try to generalize this. We have $K= \frac{x_1 + x_2 + \ldots + x_n}{n}$, and let $b_n$ be a sequence s.t $x_i=K+b_i$, and $b_1+b_2+ \ldots +b_n=0$.

So we get $P=(K+b_1)(K+b_2)\cdots(K+b_n)$

I tried to expand this, for $n=4$ for example, you get

$$ P= K^4 + K^3(x_1+x_2+x_3+x_4) + K^2 (x_1 x2 + x_1 x_3 + x_1 x_4 + \ldots) + K (x_1 x_2 x_3 + x_1 x_2 x_4 + \ldots) + x_1 x_2 x_3 x_4 $$

You can probably see the pattern, let's let $S_j$ be the sum of the products of the elements of all possible sets containing $j$ elements of all the $x_n$, we can see that

$$P=K^n+K^{n-1} S_1 + K^{n-2} S_2 + \ldots + K^{n-j} S_j+ \ldots+ KS_{n-1} + S_n$$

What is this $S_n$, and is there a formula for it? I know this has a connection with Stirling numbers and falling factorial, for $b_j=j-1$

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    $\begingroup$ Sorry, but this is illegible as written. To write math on this site, you need to use MathJax. See here for a quick intro that should tell you all you need to know. Also, to do subscripts, $a_{m,n}$ makes $a_{m,n}$. $\endgroup$ Nov 27, 2021 at 19:48
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    $\begingroup$ Could you at least parse the sentence? Also, please edit and use MathJax to properly format math expressions. $\endgroup$ Nov 27, 2021 at 19:48
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    $\begingroup$ It depends if the number $n$ is fixed or not. See here. $\endgroup$
    – Jean Marie
    Nov 27, 2021 at 19:50

1 Answer 1


These $S_n$s are called the Elementary Symmetric Polynomials and they are very well studied. Almost any question you have about them has probably been answered somewhere. Unfortunately, I don't know of a "formula" for computing $S_j$ that's better than the definition

$$S_j(x_1, \ldots, x_n) = \sum_{J \subseteq [n], |J| = j} \prod_{j \in J} x_j$$

but since it seems like you're interested in natural numbers and stirling coefficients, I do know of a closed form that might be of interest to you:

$$S_j(1,2,3,\ldots,n) = s(n,j)$$

where $s(n,j)$ are the signed stirling numbers of the first kind. See here or here for instance.

Again, much is known about the stirling numbers, so if you can phrase your question in that language, I'm sure somebody has an answer!

For instance, since you're looking for a connection to falling factorials, it's "well known" that

$$ x^{\underline{n}} = \sum_{j=0}^n s(n,k) x^k $$

so the stirling numbers act as a kind of "change of basis" from the usual polynomials $x^j$ to the falling-power polynomials $x^{\underline{j}}$.

I hope this helps ^_^

  • $\begingroup$ Can i ask another question? How can we we prove that P is maximized when all the xi's are equal using those elementary symmetric polynomials (note that we have x1+x2+...+xn=e1 (x1,x2,...,xn)=0) $\endgroup$
    – Iridium
    Nov 29, 2021 at 23:27
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    $\begingroup$ While it's normal to have some follow-up questions in the comments to answers, this seems to be a new question, even if it's related to your first one. With that in mind, you should probably ask it as a new question -- you can always leave a link to this one. I'm pretty busy right now, and I don't immediately see an answer to this follow-up question. Posing it as a new question makes it more likely for other people to see it so that you can get an answer faster ^_^ $\endgroup$ Nov 30, 2021 at 0:12

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